If you draw a line perpendicular to side c and intersecting the vertex of angle C, you make two right triangles. Let's call the perpendicular segment side d. Now the sinB = d/a, so d = a sinB = 24 * sin87° ≈ 23.967. SinA = d/b, so b = d ÷ sinA = 23.967 ÷ sin42° ≈ 36.
Let ABC and DEF be triangles which are right angled at A and D, such that the hypotenuses BC and EF are equal and, without loss of generality, angle B = angle E.ThenThen by the sine rule, BC/sin(A) = AC/sin(B) and EF/sin(D) = DF/sin(E)Since angle A = angle D = pi/2 radians, then sin(A) = sin(D) = 1so that BC/sin(A) = BC while EF/sin(D) = EFtherefore, since the hypotenuses BC and EF are equal, the left hand sides of the two equations are equal.Therefore, AC/sin(B) = DF/sin(E)then, since angle B = angle E, then sin(B) = sin(E) so that AC = DF.Also, angle C = pi/2 - angle Band angle F = pi/2 - angle Ethe right hand sides are equal so angle C = angle F.Then in a manner similar to the above, we can show that AB = DE.Thus all three pairs of corresponding sides are equal and all three pairs of corresponding angles are equal so that the two triangles are congruent.
LEG!
A circle, it has no angles.There are a infinite number of 2-d shapes that have no right angles.TrapeziumRhombusAll polygons except a square or a rectangleAll triangles except a right angle triangleAny irregular shapeetc.etc.
First identify the two lines that are parallel. Call these the top and bottom lines. Mark the two corners on the left A and B. Then angle A + angle B = 180 degrees. Next, mark the two corners on the right C and D. Angle C + angle D = 180 degrees. Hope this helps :)
If you draw a line perpendicular to side c and intersecting the vertex of angle C, you make two right triangles. Let's call the perpendicular segment side d. Now the sinB = d/a, so d = a sinB = 24 * sin87° ≈ 23.967. SinA = d/b, so b = d ÷ sinA = 23.967 ÷ sin42° ≈ 36.
Let ABC and DEF be triangles which are right angled at A and D, such that the hypotenuses BC and EF are equal and, without loss of generality, angle B = angle E.ThenThen by the sine rule, BC/sin(A) = AC/sin(B) and EF/sin(D) = DF/sin(E)Since angle A = angle D = pi/2 radians, then sin(A) = sin(D) = 1so that BC/sin(A) = BC while EF/sin(D) = EFtherefore, since the hypotenuses BC and EF are equal, the left hand sides of the two equations are equal.Therefore, AC/sin(B) = DF/sin(E)then, since angle B = angle E, then sin(B) = sin(E) so that AC = DF.Also, angle C = pi/2 - angle Band angle F = pi/2 - angle Ethe right hand sides are equal so angle C = angle F.Then in a manner similar to the above, we can show that AB = DE.Thus all three pairs of corresponding sides are equal and all three pairs of corresponding angles are equal so that the two triangles are congruent.
It can have any value in the range (0, 180) so it can be acute, right or obtuse.
LEG!
No, the pythagorean theorem only works on right triangles, but it will work on any right triangle. This is because the Pythagorean Theorem states that length of Leg A squared plus the length of Leg B Squared equals the length of the hypotenuse squared. A hypotenuse is always found opposite a right angle. Only right triangles have right angles; therefore, the Pythagorean Theorem only applies to right triangles. :D
A circle, it has no angles.There are a infinite number of 2-d shapes that have no right angles.TrapeziumRhombusAll polygons except a square or a rectangleAll triangles except a right angle triangleAny irregular shapeetc.etc.
First identify the two lines that are parallel. Call these the top and bottom lines. Mark the two corners on the left A and B. Then angle A + angle B = 180 degrees. Next, mark the two corners on the right C and D. Angle C + angle D = 180 degrees. Hope this helps :)
Any letter can be used for any angle. It is conventional, though, to go in one direction around any polygon.
Oh, dude, if ABC DEF, then congruences like angle A is congruent to angle D, angle B is congruent to angle E, and side AC is congruent to side DF would be true by CPCTC. It's like a matching game, but with triangles and math rules. So, just remember CPCTC - Corresponding Parts of Congruent Triangles are Congruent!
If the sides AB, BC and CA of triangle ABC correspond to the sides DE, EF and FD of triangle DEF, then the two triangles are congruent if:AB = DE, BC = EF and CA = FD (SSS)AB = DE, BC = EF and angle ABC = angle DEF (SAS)AB = DE, angle ABC = angle DEF, angle BCA = angle EFD (ASA)If the triangles are right angled at A and D so that BC and EF are hypotenuses, then the triangles are congruent ifBC = EF and AB = DE (RHS)BC = EF and angle ABC = angle DEF (RHA).
An 'F' angle is called a corresponding angle, a 'Z' angle is called an alternate angle and a 'C' angle is called a supplementary angle.
90 degrees of a right angle.